How do I find the standard deviation that results in a specific probability coverage in a truncated normal distribution?

Question

Given a truncated normal distribution $X$ with mean $\mu$, lower limit $a$, and upper limit $b$. How can I pick a standard deviation $\sigma$ such that $P(\mu -x\leq X \leq \mu+x)=y$ for some arbitrary $x$ and $y$.

For a standard normal distribution $N$, $$P(\mu - \sigma \leq N \leq \mu + \sigma) \approx 0.68$$ More generally, $$P(\mu - z\sigma \leq N \leq \mu + z\sigma)= erf({z \over \sqrt 2})$$ By substituting $x = z\sigma$ we get, $$P(\mu - x \leq N \leq \mu + x)= erf({x/\sigma \over \sqrt 2})$$ Then if we want the standard deviation that gives a probability of $y$ for the range $\mu - x$ to $\mu + x$ we can just solve for $\sigma$ in $$erf({x/\sigma \over \sqrt 2}) = y$$ How can we do the same for a truncated normal distribution.

Example:

Say I want a truncated normal distribution between 1 and 100 with mean 15. Additionally I want the area under the graph between 10 and 20 to be 0.7 (i.e. $cdf(20) - cdf(10) = 0.7$). How do I choose a standard deviation that fits the requirement.

Answer 1

For a truncated Normal distribution with truncation points $\underline a$ and $\overline b$, the cdf is $(\underline a\le x\le \overline b)$ $$F(X\le x)=\dfrac{\Phi(\sigma^{-1}\{x-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}$$ while the mean is $$\mu+\sigma\dfrac{\varphi(\sigma^{-1}\{\overline b-\mu\})-\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}$$ and the variance is $$\sigma^2\left[ 1+ \sigma^{-1} \dfrac{\{\overline b-\mu\}\varphi(\sigma^{-1}\{\overline b-\mu\})-\{\underline a-\mu\}\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}-\left\{\dfrac{\varphi(\sigma^{-1}\{\overline b-\mu\})-\varphi(\sigma^{-1}\{\underline a-\mu\})}{\Phi(\sigma^{-1}\{\overline b-\mu\})-\Phi(\sigma^{-1}\{\underline a-\mu\})}\right\}^2 \right]$$